TITLE : Variation In Refrigeration Coefficient Of Performance At Various
Operating Condition
1.0 INTRODUCTION
Heat flows in the direction of decreasing the temperature which is from high temperature regions to low temperature regions. This heat transfer process occurs in nature without requiring any devices. But the reverse process cannot occur by itself. The transfer of heat from a low temperature region to a high temperature one requires special devices called refrigerators. Refrigerators are cyclic devices and the working fluid used in the refrigeration is called refrigerants.
Another device that transfers heat from a low temperature medium to a high temperature one is the heat pump. Refrigerators and heat pumps are essentially the same device but differ in terms of objectives. The objective of the refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it. Discharging this heat to higher temperature medium is merely necessary part of the operation, not the purpose. The objective of a heat pump is to maintain a heated space at a high temperature. This is accomplished by absorbing heat from a low temperature source, such as well water or cold outside air in winter, and supplying this heat to a warmer medium such as a house.
Refrigeration is used widely in various applications from industrial to domestic situations, mainly for the storage and transport of perishable foodstuffs and chemical substances. It has the prime function to remove heat from a low temperature region, and it can also be applied as a heat pump for supplying heat to a region of high temperature.
2.0 OBJECTIVE
To investigate the variation of Coefficient of Performance (COPR) of a vapor compression refrigeration system at different cooling load.
3.0 THEORY
The Carnot cycle is a totally reversible cycle that consists of two reversible isothermal and two isentropic processes. It has the maximum thermal efficiency for given temperature limits and it serves as a standard against which actual power cycles can be compared. The reversed Carnot cycle is the most efficient refrigeration cycle operating between two specified temperature levels. But however, this cycle cannot be approximated in the actual devices and is not a realistic model for refrigeration cycles due to difficulty in maintaining isothermal condition during the heat absorption and heat rejection processes.
A refrigeration cycle works to lower and maintain the temperature of a controlled space by heat transfer from a low to a high temperature region.
.
Refrigeration duty is another term for the cooling effect of the refrigeration system, which is the rate of heat being removed from the low temperature region with specified evaporation and condensation temperatures. The unit for “duty” measurements is in Watts (for 1 ton of refrigeration = 3517W).
High Temperature Reservoir, TH
E
Low Temperature Reservoir, TL QH
QL
3.1 The Vapor Compression Cycle
An ideal refrigeration system follows the theoretical Reversed Carnot Cycle process. In practical refrigerators, compression and expansion of a gas and vapor mixture presents practical problems in the compressor and expander. Therefore, in practical refrigeration, compression usually takes place in the superheated condition and a throttling process is substituted for the isentropic expansion.
In an ideal vapor-compression refrigeration cycle, The refrigerant enters the condenser as superheated vapor at state 1 and leaves as saturated liquid at state 2 as a result of heat rejection to the surroundings. The temperature of the refrigerant at this state is still above the temperature of the surroundings. The saturated liquid refrigerant at state 2 is throttled to the evaporator pressure by passing it through an expansion valve or capillary tube. The temperature of the refrigerant drops below the temperature of the refrigerated space during this process.
The refrigerant enters the evaporator at state 3 as a low quality saturated mixture, and it completely evaporates by absorbing heat from the refrigerated space. The refrigerant leaves the evaporator as saturated vapor. Then the refrigerant enters the compressor at state 4 as saturated vapor and is compressed isentropically back to the condenser pressure. The temperature of the refrigerant increases during this isentropic compression process to well above the temperature of the surrounding medium.
The cycle :
1 – 2 Condensation of the high pressure vapour during which heat is transferred to the high temperature region.
2 – 3 Adiabatic throttling of the condensed vapour from the condensing to the evaporating pressure.
3 – 4 Evaporation of the low pressure liquid during which heat is absorbed from the low temperature source.
4 – 1 Isentropic compression of the vapour, from the evaporating to the condensing pressures.
Energy Transfers Analysis
Compressor
q4-1 = h4 – h1 + w4-1
If compression is adiabatic, q4-1 = 0, and w4-1 = h1 – h4 = wcomp
Power requirement, P = ṁ (h1 – h4), where ṁ is the flow rate of working fluid per unit time.
Condenser
q1-2 = h2 – h1 + w
w = 0, therefore q1-2 = h2 – h1 and rate of heat rejection Q1-2 = ṁ ( h2 – h1 )
Expansion Valve
q2-3 = h3 – h2 + w
w = 0, therefore q 2-3 = h2 – h3 and process is assumed adiabatic
q0
therefore h2 = h3Evaporator
q3–4 = h4 – h3 + w
w = 0 therefore q3–4 = h4 – h3 and rate of heat absorbed Q3–4 = ṁ ( h4 – h3 )
Coefficient of Perfomance, COPref:
COPref = w q34 = 4 1 3 4 h h h h
4.0 EQUIPMENT
5.0 PROCEDURES
6a) Condenser-water and evaporator-water
a. Water is selected as a heat source by opening valves AVS-4 and AVS-5.Then “START” is clicked
b. The water flow rate is adjusted at the condenser to 5 L/m and 3 L/m at the evaporator (evaporator load).
c. Then “COMPRESSOR” is clicked
d. When the system is stabilized, the data is recorded by click “START SAVING” e. The sampling rate is set at 120 second per sample.
f. The data for six minutes is recorded (3 samples @ 360 second). “STOP SAVING”
g. Then evaporator load is increased to 5 L/m and step (c) to step (f) is repeated.
6b) Condenser-water and evaporator-air
a. Air is selected as a heat source by opening valves AVS-3 and AVS-5.Then “START” is clicked.
b. The water flow rate is adjusted at the condenser to 5 L/m and the air flow of the evaporator is adjusted until 50% of the maximal flow (evaporator load).
c. Then “COMPRESSOR” is clicked.
d. When the system is stabilized, the data is recorded by click “START SAVING” e. The sampling rate is set at 120 second per sample.
f. The data for six minutes is recorded (3 samples @ 360 second). “STOP SAVING”
g. Then evaporator load is increased to 100% and step (c) to step (f) is repeated.
6c) Condenser-air and evaporator-air
a. Air is selected as a heat source by opening valves AVS-3 and AVS-6. Then “START” is clicked
b. The air flow of the condenser is adjusted to maximum flow (100%) and 50% of the maximal flow at the evaporator (evaporator load).
c. Then “COMPRESSOR” is clicked.
d. When the system is stabilized, the data is recorded by click “START SAVING” e. The sampling rate is set at 120 second per sample.
f. The data is recorded for six minutes (3 samples @ 360 second). “STOP SAVING”
g. Then the evaporator load is increased to 100% and step (c) to step (f) is recorded.
6d) Condenser-air and evaporator-water
a. Water is selected as a heat source by opening valves AVS-4 and AVS-6.Then “START” is clicked.
b. The air flow of the condenser is adjusted to maximum flow (100%) and the water flow rate is adjusted at the evaporator to 3 L/m (evaporator load).
c. Then “COMPRESSOR” is clicked.
d. When the system is stabilized, the data is reorded by click “START SAVING” e. The sampling rate is set at 120 second per sample.
f. The data is recorded for six minutes (3 samples @ 360 second). “STOP SAVING”
g. Then the evaporator load is increased to 5 L/m and step (c) to step (f) is repeated.
Result summary table Ex p # Tim e (s) Vref (m3/s) v2 (table) (m3/kg) mref (kg/s) = Vref/v2 h3 (kJ/kg) h4 (kJ/kg) Qevap (kW) = mref (h4 - h3) COP = Qevap/ (SW-1/1000) a 120 0.003 1.006*10-3 2.982 54.688 54.688 2451.0 7145.802 14.55 240 0.003 1.008*10-3 2.976 29.303 2470.1 7263.812 15.03 360 0.003 1.008*10-3 2.976 29.303 2470.1 7263.812 15.29 Average COP 14.96 120 0.005 1.008*10-3 4.960 `29.303 2470.1 12106.353 25.27 240 0.005 1.008*10-3 4.960 29.303 2470.1 12106.353 25.43 360 0.005 1.008*10-3 4.960 29.303 2470.1 12106.353 25.48 Average COP 25.39 b 120 0.0015 0.0199 0.075 225.94 2345.5 158.967 0.33 240 0.0015 0.0199 0.075 225.94 2357.5 159.867 0.33 360 0.0015 0.0199 0.075 225.94 2357.5 159.867 0.33 Average COP 0.33 120 0.003 0.0199 0.151 251.42 2354.5 317.565 0.66 240 0.003 0.0173 0.173 251.42 2345.5 363.832 0.75 360 0.003 0.0173 0.173 251.42 2345.5 363.832 0.75 Average COP 0.72 c 120 0.0015 0.0199 0.075 251.42 2345.5 157.056 0.35 240 0.0015 0.0199 0.075 251.42 2357.5 157.956 0.33 360 0.0015 0.0173 0.085 251.42 2357.5 179.017 0.37 Average COP 0.35 120 0.003 0.0199 0.151 251.42 2354.5 317.565 0.65 240 0.003 0.0173 0.173 251.42 2345.5 363.833 0.74 360 0.003 0.0173 0.173 251.42 2345.5 363.833 0.73 Average COP 0.71 d 120 0.003 1.008*10-3 2.976 54.688 2459.5 7156.721 14.91 240 0.003 1.010*10-3 2.970 54.688 2459.5 7156.721 14.97 360 0.003 1.010*10-3 2.970 54.688 2459.5 7156.721 14.91 Average COP 14.931 120 0.005 1.008*10-3 4.960 54.688 2459.5 7156.721 14.91 240 0.005 1.010*10-3 4.950 54.688 2459.5 7156.721 14.97 360 0.005 1.010*10-3 4.950 54.688 2459.5 7156.721 14.91 Average COP 14.93 14.931
7. RESULTS
Sample Calculations
a) Refrigerant mass flow rate,
From the experiment we can get the Volume flow rate, which is 3L/s and converted to 0.003 m/s. From the table, we can indicate the v2 which is 1/density. At 40
o
C v2 is equal to 1.006x10-3.
Mref = Vref/v2
= 0.003/1.006x10-3 = 2.982 kg/s
b) Evaporator cooling load
By using the table, we can find h3= 54.688 KJ/kg and from evaporated h4= 2451.0 KJ/kg. Qevap = Mref (H4-H3)
= 2.982 (2451.0 – 54.688) = 7145.802 KW
c) Coefficient of performance
Referring to the data obtain, SW = 491 COPref = Qevap/ SW(1-1000)
= 7145.802 / 491(1-1000) = 14.55
d) Average COPref
To get the average COPref, we need use COP from 120, 240,360s for accuracy Average COPref = (14.55+15.03+15.29)/3 = 14.96
7.0 DISCUSSIONS
From the result summary table obtained, as cooling load increase, the COPref increases. This is the effect from the mass flow rate is increase or the value of h3 decrease. When we increase the amount of cooling load definitely the mass flow rate is increase. Furthermore according from reference, a rule of thumb is that the COP improves by 2 to 4 percent for each C the evaporating temperature is raised or condensing temperature is lowered. But however, when we change the working fluid of condenser with air and the working fluid of condenser with water (experiment d) the COPref is constant. If we look closely at this state, the mass flow rate and h3 is remain constant at any time and so does h4, that is why the COPref also does not change and remain constant. Along the experiment was done, there might have some error such as before saving, the system maybe does not 100% stabilized. The graph shown on the screen which shows the stabilization cannot be 100% accurate as same as the actual system.
We look forward to the cooling medium effect, there is two types of medium used which is water (experiment a) and air (experiment b). When water is used as the cooling medium, the COP is much higher compare to air. The higher the COP equate to lower operating cost. This shows that liquid/water cooling is a better cooling medium. This is becauseair is not an outstanding thermal conductor (air has a thermal conductivity of 0.026 W/mK). A more effective method of cooling than air cooling is liquid/water cooling as the heat transferring medium. One way of increasing the thermal efficiency of a water cooling system is by placing coils inside the cooling channel to induce a turbulent flow of the cooling liquid.
Load is the amount of heat energy to be removed from refrigerators by the HVAC equipment to maintain the design temperature. In domestic fridge, the refrigeration cycle for the fresh food compartment could be used directly subcool the condensate for the freezer cycle, thereby shifting some of the cooling load from the freezer to the fresh food cycle. Comparing the COP of a dual-cycle system to single-cycle system, in a related study found a 23% improvement in overall COP for a dual cycle system using refrigerants R-142 and R-152 when the total cabinet loads are evenly distributed between the freezer and the fresh food compartment.
Figure 1 : Refrigeration System Operating Characteristics
The refrigeration practice (with actual loads) in a factory. The refrigeration process begins with the compressor. The process start when Ammonia gas is compressed until it becomes very hot from the increased pressure. This heated gas flows through the coils behind the refrigerator, which allow excess heat to be released into the surrounding air. This is why users sometimes feel warm air circulating around the fridge. Eventually the ammonia cools down to the point where it becomes a liquid. This liquid form of ammonia is then forced through a device called an expansion valve. Essentially, the expansion valve has such a small opening that the liquid ammonia is turned into a very cold, fast-moving mist, evaporating as it travels through the coils in the freezer. Since this evaporation occurs at -27 degrees F (-32 degrees Celsius), the ammonia draws heat from the surrounding area. This is the Second Law of Thermodynamics in effect. Cold material, such as the evaporating ammonia gas, tends to take heat from warmer materials, such as the water in the ice cube tray.
As the evaporating ammonia gas absorbs more heat, its temperature rises. Coils surrounding the lower refrigerator compartment are not as compact. The cool ammonia still draws heat from the warmer objects in the fridge, but not as much as the freezer section. The ammonia gas is drawn back into the compressor, where the entire cycle of pressurization, cooling and evaporation begins anew
Figure 2: Batch-continuous air blast freezer with counter flow air circulation
Figure 3: Batch continuous air blast freezer with cross flow
8.0 CONCLUSION
As the conclusion, refrigerator consists with two compartments - one for frozen items and the other for items requiring refrigeration but not freezing. The objective of this experiment have been completely achieved by understanding the relation of various condition with Coefficient of Performance, all the parameters required to be solved have been calculated and solved accordingly. In addition, all of the experiments have eventually being done according to the procedures given systematically and appropriately.
9.0 REFERENCES
1) Cengel, Ghajar., Heat and Mass Transfer : Fundamentals and Applications, 4th edition in SI Units.
2) Cengel, Boles .,Thermodynamics : An Engineering Approach, 7th edition in SI units
3) UiTM, Faculty of Mechanical Engineering ., MEC551 Thermal Engineering, 2013.
4) Welty, James R., et al. Fundamentals of momentum, heat, and mass transfer. John Wiley & sons, 2009.
5) http://www.fridgesolutions.com/terms-considerations.html. Retrieved 4 April 2014
